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Proof step 19 from proof subset-size (previous step)

r -- sk ∈ st0, r ∈ st0, r ∈ st0 .refs, r ∈ st1 .refs, # st0 .refs < # st1 .refs, st1 ⊆ st0 -- (r -- sk), ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st1 = ∅, st0 = ∅, r -- sk ∈ st1, ∀ rk. rk ∈ st1 → rk ∈ st0 ∧ st1[rk] = st0[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
st1 ⊆ st0 -- (r -- sk), # st0 .refs < # st1 .refs, r ∈ st1 .refs, r ∈ st0 .refs, r ∈ st0, r -- sk ∈ st0, rk ∈ st1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, st1 = ∅, r -- sk ∈ st1, rk ∈ st0 ∧ st1[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]